Classical Physics - Pulley Problem

A massless string is placed over a massless pulley, and each end is wound around and fastened to a vertical hoop. The hoops have masses M1 and M2 and radii R1 and R2. The apparatus is placed in a uniform gravitation field g and released with each end of the string aligned along the field.

I have to show that the tension is T = gM1M2/(M1+M2)

I can sort of solve the problem by just saying that (M1M2)/(M1+M2) is the reduced mass. Then all we have to do is say the tension is balanced with the weight, so that T = Mg = gM1M2/(M1+M2). But then doesn't this imply that the hoop isn't moving? I think I'm missing something here... It is also important to realize that both hoops are rolling down the string so the acceleration of one hoop downward is not necessarily the acceleration of the other one upward... Any help / hints would be appreciated... thanks

Try this. Each hoop has a torque on it of T times the appropriate radius. This determines the angular acceleration alpha of each hoop. Let s be the amount of string as a function of time. Then I claim s''=a1+a2=R1*alpha1+R2*alpha2 where the a's are the linear acceleration of each hoop. (The sum of the center of mass accelerations determines the second derivative of the length of string but the amount is also determined by the angular accelerations). Put this together with the usual force balance diagram for the vertical forces on each hoop and you will get what you want.

Can you work out a1 and a2 from linear force balance? Can you work out alpha1 and alpha2 from moment of inertia and torque? Let's get started... Do you accept the equality of the two expressions for s''? Does it seem right?